SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Now, let  $X = N \cup S$ , i.e., we are in the last iteration of the algorithm. Due to an analogous reasoning as before, there is a many-to-one mapping from the min-cost branchings in  $G$  to branchings in  $G_{X}$ . More precisely, for every branching  $B \in \mathcal{B}_{Y,\{s\}}(G_X)$ , there exist
$$
\prod_ {(Y, Y ^ {\prime}) \in B} w _ {X} (Y, Y ^ {\prime}) = w _ {X} (B)
$$
branchings in  $G$  that map to  $B$ . Hence, by the Markov chain tree theorem (Lemma 3), we get
$$
A _ {v, s} = Q _ {v, s} = \frac {\sum_ {B \in \mathcal {B} _ {Y _ {v} , \{s \}} (G _ {X})} w _ {X} (B)}{\sum_ {B \in \mathcal {B} (G _ {X})} w _ {X} (B)} = \frac {\sum_ {B \in \mathcal {B} _ {v , s} ^ {*} (G)} 1}{\sum_ {B \in \mathcal {B} ^ {*} (G)} 1},
$$
where  $(G_X', P)$  is the Markov chain corresponding to  $G_X$  and  $Q = \lim_{\tau \to \infty} \frac{1}{\tau} \sum_{i=0}^{\tau} P^{\tau}$ . This equals the definition of MIXED BORDA BRANCHING.
Lastly, we argue about the running time of the algorithm. For a given delegation graph  $(G,c)$ , let  $n = V(G)$ , i.e., the number of voters. Algorithm 1 can be implemented in  $\mathcal{O}(n^3)$ . That is because, the while loop runs for  $\mathcal{O}(n)$  iterations (the laminar set family  $\mathcal{F}$  can have at most  $2n - 1$  elements), and finding all strongly connected components in a graph can be done in  $\mathcal{O}(n^2)$  (e.g., with Kosaraju's algorithm [Hopcroft et al., 1983]). Coming back to the running time of Algorithm 2, we note that the do-while loop runs for  $\mathcal{O}(n)$  iterations, again, due to the size of  $\mathcal{F}'$ . In line 7, the algorithm computes  $\mathcal{O}(n)$  times the number of weighted spanning trees with the help of Lemma 1 (Tutte [1948]). Hence, the task is reduced to calculating the determinant of a submatrix of the laplacian matrix. Computing an integer determinant can be done in polynomial time in  $n$  and  $\log(M)$ , if  $M$  is an upper bound of all absolute values of the matrix<sup>12</sup>. Note, that all values in every Laplacian (the out-degrees on the diagonals and the multiplicities in the other entries) as well as the results of the computation are upper-bounded by the total number of branchings in the original graph  $G$  (this follows from our argumentation about the interpretation of  $t_X(v)$  in the proof of Theorem 4), hence in particular by  $n^n$ . Therefore, the running time of each iteration of the do-while loop is polynomial in  $n$ . In the final step we compute the absorbing probabilities of the (scaled down version) of the weighted graph  $(G_X,w_X)$  (where  $X = N \cup S$ ). For that, we need to compute the inverse of a  $\mathcal{O}(n) \times \mathcal{O}(n)$  matrix, which can be done using the determinant and the adjugate of the matrix. Computing these comes down to computing  $\mathcal{O}(n^2)$  determinants, for which we argued before that it is possible in polynomial time<sup>13</sup>. Summarizing, this gives us a running time of Algorithm 1 in  $\mathcal{O}((n^7\log(n) + n^4\log(n\log(n))) * (\log^2 n + (\log(n\log(n)))^2))$ .
B Further Remarks on Section 6
Alternative Interpretation of Algorithm 2 We stated Algorithm 2 in terms of counting min-cost branchings. There exists a second natural interpretation that is closer to the definition of the RANDOM WALK RULE, in which we want to compute the limit of the absorbing probabilities of a parametric Markov chain. We give some intuition on this reinterpretation of the algorithm with the example in Figure 2, and later extend this interpretation to a larger class of parametric Markov chains.
Intuitively speaking, every set  $X \in \mathcal{F}$  in the Markov chain  $(G, P^{(\varepsilon)})$  corresponding to the delegation graph  $G$  is a strongly connected component whose outgoing edges have an infinitely lower probability than the edges inside of  $X$  as  $\varepsilon$  approaches zero. We are therefore interested in the behavior of an infinite random walk in  $G[X]$ . While in the branching interpretation, the node weight  $t_X(v)$  can be interpreted as the number of min-cost  $v$ -arborescences in  $G[X]$ , in the Markov chain interpretation we think of  $t_X(v)$  as an indicator of the relative time an infinite random walk spends in  $v$  (or the relative number of times  $v$  is visited) in the Markov chain given by the strongly connected graph  $G[X]$ . Consider the example iteration depicted in Figure 2a, where we are given an unprocessed  $X \in \mathcal{F}$  whose children  $Y_1, Y_2$  are all processed. When contracting  $Y_1$  and  $Y_2$  the weights on the edges should encode how likely a transition is from one set to another, which is achieved by summing over the relative time spent in each node with a corresponding edge. We then translate the resulting
12More precisely, it can be computed in  $\mathcal{O}((n^4\log (nM) + n^3\log^2 (nM))*(\log^2 n + (\log \log M)^2))$  [Gathen and Gerhard, 2013]
<sup>13</sup>We argued this only for integer matrices, but we can transform the rational matrix into an integer one by scaling it up by a factor which is bounded by  $n^n$ .
graph (Figure 2b) into a Markov chain and again compute the relative time spend in each node. This computation is equivalent to calculating the sum of weights of all in-trees (up to a scaling factor, see Theorem 3). Indeed, we get a ratio of one to three for the time spend in  $Y_{1}$  and  $Y_{2}$ . To compute  $t_{X}(v)$  we multiply the known weight  $t_{Y_v}(v)$  by the newly calculated weight of  $Y_{v}$ . In the example this means that since we know, we spend three times as much time in  $Y_{2}$  as in  $Y_{1}$  all weights of nodes in  $Y_{2}$  should be multiplied by three (see Figure 2c).
Extension of Algorithm 2 In addition, we remark that our algorithm could be extended to a larger class of parametric Markov chains, namely, to all Markov chains  $(G,P^{(\varepsilon)})$ , where  $G$  is a graph in which every node has a path to some sink node, and, for every  $e\in E(G)$ ,  $P_{e}^{(\varepsilon)}$  is some rational fraction in  $\varepsilon$ , i.e.,  $\frac{f_e(\varepsilon)}{g_e(\varepsilon)}$ , where both  $f_{e}$  and  $g_{e}$  are polynomials in  $\varepsilon$  with positive coefficients.14 Now, we can construct a cost function  $c$  on  $G$ , by setting  $c(e) = x_{e} - z_{e} + 1$ , where  $x_{e}$  is the smallest exponent in  $f_{e}(\varepsilon)$  and  $z_{e}$  is the smallest exponent in  $g_{e}(\varepsilon)$ . Note that, if  $c(e) < 1$ , then the Markov chain cannot be well defined for all  $\varepsilon \in (0,1]$ . Now, we run Algorithm 2 for the delegation graph  $(G,c)$  with the only one difference, i.e., the weight function  $w_{X}$  also has to incorporate the coefficients of the polynomials  $f_{e}(\varepsilon)$  and  $g_{e}(\varepsilon)$ . More precisely, we define for every  $e\in E$ , the number  $q_{e}$  as the ratio between the coefficient corresponding to the smallest exponent in  $f_{e}$  and the coefficient corresponding to the smallest exponent in  $g_{e}$ . Then, we redefine line 4 in the algorithm to be
$$
w_{X}(Y,Y^{\prime})\gets \sum_{(u,v)\in E_{y}\cap (Y\times Y^{\prime})}t_{Y}(u)\cdot q_{(u,v)}.
$$
One can then verify with the same techniques as in Section 5 and Section 6, that this algorithm returns the outcome of the above defined class of Markov chains.
C Missing Proofs and Further Results of Section 7
Proof. Given a delegation graph  $(G, c)$  and a bijection  $\sigma: V(G) \to V(G)$ , we know that for all  $v \in V(G)$  it holds that  $|\delta_G^+(v)| = |\delta_{G'}^+(\sigma(v))|$  and  $c(v, w) = c'(\sigma(v), \sigma(w))$  for any edge  $(v, w) \in \delta^+(v)$ , where  $(G', c') = \sigma((G, c))$ . In the corresponding Markov chains  $M_{\varepsilon}$  and  $M_{\varepsilon}'$  we therefore get  $P_{(v, w)}^{(\varepsilon)} = P_{(\sigma(v), \sigma(w))}^{(\varepsilon)}$  (see Equation 1). Since through the bijection between the edges of the graph, we also get a bijection between all walks in the graph  $\mathcal{W}$  and for every  $s \in S$  and walk in  $\mathcal{W}[s, v]$  there is a corresponding walk in  $\mathcal{W}[\sigma(v), \sigma(s)]$  of the same probability. Therefore we have
$$
A _ {v, s} = \lim _ {\varepsilon \to 0} \sum_ {W \in \mathcal {W} [ v, s ]} \prod_ {e \in W} P _ {e} ^ {(\varepsilon)} = \lim _ {\varepsilon \to 0} \sum_ {W \in \mathcal {W} [ \sigma (v), \sigma (s) ]} \prod_ {e \in W} P _ {e} ^ {\prime (\varepsilon)} = A _ {\sigma (v), \sigma (s)} ^ {\prime},
$$
Theorem 7 ( $\star$ ). The RANDOM WALK RULE satisfies copy-robustness.
Proof. Let  $(G, c), v, (\hat{G}, c), A, \hat{A}$  and  $S_v$  be defined as in the definition of copy-robustness. Let  $(\mathcal{F}, y)$  and  $(\hat{\mathcal{F}}, \hat{y})$  be the set families and functions returned by Algorithm 1 for  $G$  and  $\hat{G}$ , respectively. In this proof, we restrict our view to the subgraphs of only tight edges, denoted by  $G_y = (N \cup S, E_y)$  and  $\hat{G}_{\hat{y}} = (N \setminus \{v\} \cup V \cup \{v\}, E_{\hat{y}})$ , respectively. Note, that this does not change the result of the RANDOM WALK RULE, since it is shown to be equal to MIXED BORDA BRANCHING, which only considers tight edges (in the contracted graph) itself.
First, we observe that the set  $S_{v}$  is exactly the subset of  $S$  reachable by  $v$  in  $G_{y}$ . This is because the assignment  $A$  returned by the RANDOM WALK RULE is given as the absorbing probability of a Markov chain on the graph  $(G_{X}, w_{X})$  with  $X = N \cup S$ , computed by Algorithm 2. The graph is constructed from  $G_{y}$  by a number of contractions, which do not alter reachability, i.e. for  $s \in S$  the node  $\{s\}$  is reachable from the node  $Y_{v}$  containing  $v$  in  $G_{X}$  exactly if  $s$  is reachable from  $v$  in  $G_{y}$ . Since all edge weights  $w_{X}$  are strictly positive, in the corresponding Markov chain all transition
This class is reminiscent of a class of parametric Markov chains studied by Hahn et al. [2011].
probabilities on the edges of  $G_{X}$  are strictly positive as well. This gives  $\{s\}$  a strictly positive absorbing probability when starting a random walk in  $Y_{v}$  exactly if  $s$  is reachable from  $v$  in  $G_{y}$ .
Our next observation is that  $\hat{\mathcal{F}} = \mathcal{F} \setminus \{Y \in \mathcal{F} \mid v \in Y\} \cup \{\{v\}\}$ ,  $\hat{y}(\{v\}) = 1$  and  $y(Y) = \hat{y}(Y)$  for all  $Y \in \hat{\mathcal{F}} \setminus \{\{v\}\}$ . Consider the computation of  $\mathcal{F}$  in Algorithm 1. Since the output is unique (see Lemma 2 statement (i)), we can assume without loss of generality that after initializing  $\mathcal{F}$ , all sets in  $\{Y \in \mathcal{F} \mid v \notin Y\}$  are added to  $\mathcal{F}$  first and then the remaining sets  $\{Y \in \mathcal{F} \mid v \in Y\}$ . In  $\hat{G}$ , the only edges missing are the outgoing edges from  $v$ , therefore, when applying Algorithm 1 to  $\hat{G}$  all sets in  $\{Y \in \mathcal{F} \mid v \notin Y\}$  can be added to  $\hat{\mathcal{F}}$  first (with  $\hat{y}(Y) = y(Y)$ ). Note, that the set  $\{v\}$  with  $y(\{v\}) = 1$  was added to  $\hat{\mathcal{F}}$  in the initialization. We claim, that the algorithm terminates at that point. Suppose not, then there must be another strongly connected component  $X \subseteq N$  with  $\delta^{+}(X) \cap E_{\hat{y}} = \emptyset$ . If  $v \in X$  then since  $v$  has no outgoing edges  $X = \{v\}$ , which is already in  $\mathcal{F}$ . If  $v \notin X$  then  $X$  would have already been added.
With these two observations, we can show the following claim: For every casting voter  $s \in S \setminus S_v$  the voting weight remains equal, when  $v$  turns into a casting voter, i.e.,  $\pi_s(A) = \pi_s(\hat{A})$ . Fix  $s \in S \setminus S_v$  and let  $U \subset N$  be the set of nodes not reachable from  $v$  in  $G_y$ . We know that  $\hat{\mathcal{F}} = \mathcal{F} \setminus \{Y \in \mathcal{F} \mid v \in Y\} \cup \{v\}$ , which implies that for every node  $u \in U$  the sets containing  $u$  are equal in  $\mathcal{F}$  and  $\hat{\mathcal{F}}$ , i.e.,  $\{Y \in \mathcal{F} \mid u \in Y\} = \{Y \in \hat{\mathcal{F}} \mid u \in Y\}$ . Therefore, the outgoing edges from any  $u \in U$  are equal in  $G_y$  and  $\hat{G}_{\hat{y}}$ . Since  $\hat{\mathcal{F}} \subseteq \mathcal{F}$ , the edges in  $\hat{G}_{\hat{y}}$  are a subset of the edges in  $G_y$  and therefore the set  $U$  is not reachable from  $v$  in  $\hat{G}_{\hat{y}}$ . When translating  $\hat{G}_{\hat{y}}$  into the Markov chain  $(\hat{G}_{\hat{y}}, \hat{P}^{(\varepsilon)})$  (see Equation 1), we get for the probability of any tight out-edge  $e$  of  $u$  and any  $\varepsilon > 0$ , that  $P_e^{(\varepsilon)} = \hat{P}_e^{(\varepsilon)}$ , where  $P^{(\varepsilon)}$  is the transition matrix induced by the original graph  $G_y$ . In the following we argue about the set of walks in  $G_y$  and  $G_{\hat{y}}$ . To this end we define for every  $u \in N$ , the set  $\mathcal{W}[u, s]$  ( $\hat{\mathcal{W}}[u, s]$ , respectively) as the set of walks in  $G_y$  (in  $G_{\hat{y}}$ , respectively) that start in  $u$  and end in sink  $s$ . Since all walks from any  $u \in U$  to  $s$  contain only outgoing edges from nodes in  $U$ , we have  $\hat{\mathcal{W}}[u, s] = \mathcal{W}[u, s]$ . For any other voter  $w \in N \setminus U$  we have  $\hat{\mathcal{W}}[w, s] = \mathcal{W}[w, s] = \emptyset$  and therefore
$$
\pi_ {s} (\hat {A}) = 1 + \sum_ {u \in U} \lim _ {\varepsilon \to 0} \sum_ {\hat {W} \in \hat {\mathcal {W}} [ u, s ]} \prod_ {e \in \hat {W}} P _ {e} ^ {(\varepsilon)} = 1 + \sum_ {u \in U} \lim _ {\varepsilon \to 0} \sum_ {W \in \mathcal {W} [ u, s ]} \prod_ {e \in W} P _ {e} ^ {(\varepsilon)} = \pi_ {s} (A) ,
$$
which concludes the proof of the claim.
Summarizing, we know that for any casting voter  $s \in S \setminus S_v$  we have  $\pi_s(A) = \pi_s(\hat{A})$ , which directly implies that  $\sum_{s \in S_v} \pi_s(A) = \pi_v(\hat{A}) + \sum_{s \in S_v} \pi_s(\hat{A})$ .
Proof. Before proving the claim, we introduce notation. For any walk  $W$  in some graph  $G$ , and some node  $v \in V(G)$ , we define  $W[v]$  to be the subwalk of  $W$  that starts at the first occasion of  $v$  in  $W$ . For two nodes  $u, v \in V(G)$ , we define  $W[u, v]$  to be the subwalk of  $W$  that starts at the first occasion of  $u$  and ends at the first occasion of  $v$ . (Note that  $W[v]$  and  $W[u, v]$  might be empty.) Now, for a set of walks  $\mathcal{W}$  and  $u, v, s \in V(G)$ , we define  $\mathcal{W}[v] = \{W[v] \mid W \in \mathcal{W}\}$  and  $\mathcal{W}[u, v] = \{W[u, v] \mid W \in \mathcal{W}\}$ . Lastly, we define  $\mathcal{W}[u, v, s] = \{W \in W[u, s] \mid v \in W[u, s]\}$ . We usually interpret a walk  $W$  as a sequence of nodes. In order to facilitate notation, we abuse notation and write  $v \in W$  for some node  $v \in V(G)$  in order to indicate that  $v$  appears in  $W$ , and for an edge  $e \in E(G)$ , we write  $e \in W$  to indicate that tail and head of  $e$  appear consecutively in  $W$ .
For the remainder of the proof we fix  $\mathcal{W}$  to be the set of walks in the input delegation graph  $G$  starting in some node from  $N$  and ending in some sink node  $S$ . Moreover, let  $G_{X}$  be the graph at the end of Algorithm 2, i.e.,  $G_{X}$  for  $X = N \cup S$ . We fix  $\hat{\mathcal{W}}$  to be the set of walks which start in some node of  $G_{X}$  and end in some sink node of  $G_{X}$  (which are exactly the nodes in  $\{\{s\} \mid s \in S\}$ ).
In the following, we define for every  $v \in N$  a probability distribution  $f_v: \mathcal{W}[v] \to [0,1]$ , such that it witnesses the fact that the RANDOM WALK RULE is confluent. To this end, we define a mapping  $\gamma_v: \hat{\mathcal{W}}[Y_v] \to \mathcal{W}[v]$ , where  $Y_v$  is the node in  $G_X$  that contains  $v$ . Given a walk  $\hat{W} \in \hat{\mathcal{W}}[Y_v]$ , we construct  $\gamma_v(\hat{W}) \in \mathcal{W}[v]$  as follows: Let  $\hat{W} = Y^{(1)}, \ldots, Y^{(k)}$ . By construction of  $G_X$  we know that for every  $i \in \{1, \ldots, k\}$ , the fact that  $(Y^{(i)}, Y^{(i+1)}) \in E_X$  implies that there exists  $(b^{(i)}, a^{(i+1)}) \in E$  with  $b^{(i)} \in Y^{(i)}$  and  $a^{(i+1)} \in Y^{(i+1)}$ . Moreover, we define  $a^{(1)} = v$  and  $b^{(n)} = s$ ,
where  $\{s\} = Y^{(k)}$ . Under this construction it holds that  $a^{(i)}, b^{(i)} \in Y^{(i)}$  for all  $i \in \{1, \dots, k\}$ , but the two nodes may differ. Therefore, we insert subwalks  $W^{(i)}$  connecting  $a^{(i)}$  to  $b^{(i)}$  by using only nodes in  $Y^{(i)}$  and visiting each of these nodes at least once. The final walk  $\gamma_v(\hat{W})$  is then defined by  $(a^{(1)}, W^{(1)}, b^{(1)}, \dots, a^{(n)}, W^{(n)}, b^{(n)})$ . We remark that this mapping is injective, and it holds that  $\hat{W}$  visits some node  $Y \in V(G_X)$  if and only if  $\gamma_v(\hat{W})$  visits all nodes in  $Y$ .
Recall that the assignment  $A$  of the RANDOM WALK RULE can be computed via a Markov chain  $(G_X', P)$  derived from the contracted graph  $(G_X, w_X)$  (see Section 5 and Section 6), where  $G_X'$  is derived from  $G_X$  by adding self-loops. In Lemma A.1 we show that introducing (and thus removing) self-loops to states in an absorbing Markov chain does not change its absorbing probabilities. We retrieve the Markov chain  $(G_X, \hat{P})$  by removing all self loops of all voters in  $N$  and rescaling the other probabilities accordingly. We then make use of this Markov chain in order to define  $f_v$  over  $\mathcal{W}[v]$ . That is, for any  $W \in \mathcal{W}[v]$  let
$$
f _ {v} (W) = \left\{ \begin{array}{l l} \prod_ {e \in \hat {W}} \hat {P} _ {e} & \text {i f t h e r e e x i s t s} \hat {W} \in \hat {\mathcal {W}} [ Y _ {v} ] \text {s u c h t h a t} \gamma_ {v} (\hat {W}) = W \\ 0 & \text {e l s e .} \end{array} \right.
$$
Note that, the above expression is well defined since  $\gamma_v$  is injective.
In the remainder of the proof, we show that  $f_v$  witnesses the confluence of the RANDOM WALK RULE. First, we show that  $f_v$  is indeed consistent with the assignment  $A$  returned by RANDOM WALK RULE. That is, for any  $v \in N$  and  $s \in S$  it holds that
$$
\mathbb {P} _ {W \sim f _ {v}} [ s \in W ] = \sum_ {W \in \mathcal {W} [ v, s ]} f _ {v} (W) = \sum_ {\hat {W} \in \hat {\mathcal {W}} \left[ Y _ {v}, \{s \} \right]} \prod_ {e \in \hat {W}} \hat {P} _ {e} = A _ {v, s} \quad .
$$
The second equality comes from the fact that  $\gamma_v$  is injective and exactly those walks in  $\hat{\mathcal{W}}[Y_v, \{s\}]$  are mapped by  $\gamma_v$  to walks in  $\mathcal{W}[v, s]$ . Moreover, all walks in  $\mathcal{W}[v, s]$  that have no preimage in  $\hat{\mathcal{W}}[Y_v, \{s\}]$  are zero-valued by  $f_v$ . The last equality comes from the fact that  $A_{v,s}$  equals the probability that the Markov chain  $(G_X', P)$  (equivalently,  $(G_X, \hat{P})$ ) reaches  $\{s\}$  if started in  $Y_v$  (see Section 5 and Section 6).
We now turn to the second condition on the family of probability distributions  $f_v, v \in N$ . That is, for every  $u, v \in N, s \in S$  it holds that
$$
\begin{array}{l} \mathbb {P} _ {W \sim f _ {u}} [ v \in W \land s \in W ] = \sum_ {W \in \mathcal {W} [ u, v, s ]} f _ {u} (W) = \sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {u}, Y _ {v}, \{s \} ]} \prod_ {e \in \hat {W}} \hat {P} _ {e} \\ = \sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {u}, Y _ {v}, \{s \} ]} \left(\prod_ {e \in \hat {W} [ Y _ {u}, Y _ {v} ]} \hat {P} _ {e}\right) \left(\prod_ {e \in \hat {W} [ Y _ {v}, \{s \} ]} \hat {P} _ {e}\right) \\ = \big (\sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {u}, Y _ {v} ]} \prod_ {e \in \hat {W}} \hat {P} _ {e} \big) \cdot \big (\sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {v}, \{s \} ]} \prod_ {e \in \hat {W}} \hat {P} _ {e} \big) \\ = \left(\sum_ {s ^ {\prime} \in S} \sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {u}, Y _ {v}, \{s ^ {\prime} \} ]} \prod_ {e \in \hat {W}} \hat {P} _ {e}\right) \cdot \left(\sum_ {\hat {W} \in \hat {\mathcal {W}} [ Y _ {v}, \{s \} ]} \prod_ {e \in \hat {W}} \hat {P} _ {e}\right) \\ = \left(\sum_ {s ^ {\prime} \in S} \sum_ {W \in \mathcal {W} [ u, v, s ^ {\prime} ]} f _ {u} (W) \cdot \left(\sum_ {W \in \mathcal {W} [ v, s ]} f _ {v} (W)\right) \right. \\ = \mathbb {P} _ {W \sim f _ {u}} [ v \in W ] \cdot \mathbb {P} _ {W \sim f _ {v}} [ s \in W ]. \\ \end{array}
$$
The second equality follows from the same reason as above, i.e.,  $\gamma_v$  is injective, exactly those walks in  $\hat{\mathcal{W}}[Y_u, Y_v, \{s\}]$  are mapped by  $\gamma_v$  to walks in  $\mathcal{W}[u, v, s]$ , and all walks in  $\mathcal{W}[u, v, s]$  that have no preimage in  $\hat{\mathcal{W}}[Y_u, Y_v, \{s\}]$  are zero-valued by  $f_v$ . The third inequality holds by the fact that every walk that is considered in the sum can be partitioned into  $\hat{W}[Y_u, Y_v]$  and  $\hat{W}[Y_v, \{s\}]$ . The fourth equality follows from factoring out by the subwalks. The fifth equality follows from the fact that every walk in  $\hat{\mathcal{W}}$  reaches some sink node eventually, and therefore, the additional factor in the first bracket sums up to one. Lastly, the sixth equality follows from the very same argument as before.
From the above equation we get in particular that for every  $u, v \in N$ ,  $s \in S$  it holds that
$$
\mathbb {P} _ {W \sim f _ {u}} [ s \in W | v \in W ] = \frac {\mathbb {P} _ {W \sim f _ {u}} [ s \in W \land v \in W ]}{\mathbb {P} _ {W \sim f _ {u}} [ v \in W ]} = \mathbb {P} _ {W \sim f _ {v}} [ s \in W ].
$$
The next axiom was in its essence first introduced by Behrens and Swierczek [2015] and first given the name guru-participation in Kotsialou and Riley [2020]. The idea is that a representative (the guru) of a voter, should not be worse off if said voter abstains from the vote. Brill et al. [2022] define this property for non-fractional ranked delegations by requiring that any casting voter that was not a representative of the newly abstaining voter should not loose voting weight. This definition translates well into the setting of fractional delegations where we can have multiple representatives per voter. For simplicity, we made a slight modification to the definition<sup>15</sup>, resulting in a slightly stronger axiom.
Previously, we stated the general assumption that every delegating voter in a delegation graph  $(G,c)$  has a path to some casting voter in  $G$ . In this section we modify given delegation graphs by removing nodes or edges, which may result in an invalid delegation graph not satisfying this assumption. To prevent this, we implicitly assume that after modifying a delegation graph, all nodes in  $N$  not connected to any sink in  $S$  (we call them isolated) are removed from the graph.
Guru Participation: A delegation rule satisfies guru-participation if the following holds for every instance  $(G, c)$ : Let  $(\hat{G}, c)$  be the instance derived from  $(G, c)$  by removing a node  $v \in N$  (and all newly isolated nodes), let  $S_v = \{s \in S \mid A_{v,s} > 0\}$  be the set of representatives of  $v$  and let  $A$  and  $\hat{A}$  be the assignments returned by the delegation rule for  $(G, c)$  and  $(\hat{G}, c)$ , respectively. Then
$$
\pi_ {s} (\hat {A}) \geq \pi_ {s} (A) \quad \forall s \in S \backslash S _ {v}.
$$
In particular, this implies
$$
\sum_ {s \in S _ {v}} \pi_ {s} (\hat {A}) + 1 \leq \sum_ {s \in S _ {v}} \pi_ {s} (A) \quad .
$$
In order to prove that the RANDOM WALK RULE satisfies guru-participation we first show the following lemma, saying that the voting weight of no casting voter decreases, when the in-edges of another casting voter are removed from the graph.
Lemma C.1. For the RANDOM WALK RULE, removing the incoming edges of some casting voter  $s \in S$  (and all newly isolated voters) does not decrease the absolute voting weight of any casting voter  $s' \in S \setminus \{s\}$ .